Problem: Find the gradient of $f(x, y) = x^2 - xy$ at $(1, -1)$. $\nabla f(1, -1) = ($
The gradient of a scalar field is all its partial derivatives put together into a vector. For a 2D scalar field, this looks like $\nabla f = (f_x, f_y)$. Let's find $f_x$ and $f_y$. $\begin{aligned} f_x &= \dfrac{\partial}{\partial x} \left[ x^2 - xy \right] \\ \\ &= 2x - y \\ \\ f_y &= \dfrac{\partial}{\partial y} \left[ x^2 - xy \right] \\ \\ &= -x \end{aligned}$ Now we can evaluate the partial derivatives we found at the point $(1, -1)$. $\begin{aligned} f_x(1, -1) &= 2x - y = 2 - (-1) = 3 \\ \\ f_y(1, -1) &= -x = -1 \end{aligned}$ The gradient of $f$ at $(1, -1)$ is $\nabla f = (3, -1)$.